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suss
 2 years ago
Xlogylogz.ylogzlogx.zlogxlogy=1
how to prove this??
suss
 2 years ago
Xlogylogz.ylogzlogx.zlogxlogy=1 how to prove this??

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suss
 2 years ago
Best ResponseYou've already chosen the best response.0it was supposed to x to the power logylogz y to the power logzlogx z to the power logxlogy

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1\(\huge x^{\log y\log z}y^{\log z\log x}z^{\log x\log y}\) like this ?

suss
 2 years ago
Best ResponseYou've already chosen the best response.0ah thank god..yes like that

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1consider \(\huge \log[x^{\log y\log z}y^{\log z\log x}z^{\log x\log y}]\) can u simplify that ?

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1use, \(\log AB=\log A+\log B \\ \log A^{B}=B\log A\)

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1you should get that =0 = ln 1

suss
 2 years ago
Best ResponseYou've already chosen the best response.0i dont understand y u multiplied the whole by log????

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1else, nothing can't be really done to that expression

suss
 2 years ago
Best ResponseYou've already chosen the best response.0i have done that problem in this way \[x ^{\log y/z}y ^{\log z/x}z ^{\log x/y}\]

suss
 2 years ago
Best ResponseYou've already chosen the best response.0and i cudnt figure out the next step..... :(

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1i know, i tried that also....since i could not proceed, i thought of taking log....

suss
 2 years ago
Best ResponseYou've already chosen the best response.0if u dont mind,can u show me the simplification process...??? :)

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1\(\huge \log[x^{\log y\log z}y^{\log z\log x}z^{\log x\log y}] \\\large =\log[x^{\log y\log z}]+\log[y^{\log z\log x}]+\log[z^{\log x\log y}]\) got this ?

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1now use \( \\ \log A^{B}=B\log A\)

suss
 2 years ago
Best ResponseYou've already chosen the best response.0oh sorry man i jus lost connection leeme check it out

suss
 2 years ago
Best ResponseYou've already chosen the best response.0y did u keep the + sign ???

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1because \(\log AB=\log A+\log B \\ \)

suss
 2 years ago
Best ResponseYou've already chosen the best response.0ah finally got it .....thx a ton man

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1sure ? u getting that = 0 ? then write 0=log 1

suss
 2 years ago
Best ResponseYou've already chosen the best response.0yes 100%..i hav to use formula of log ab bak ther ......i really got it man
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